Convergence of $u_{k+1} = \frac{1}{2-u_k u_{k-1}}$ Let $u_0, u_1 \in \mathbb{R}$ be arbitrary. I am interested in the sequence defined by:
$$u_{k+1} = \frac{1}{2-u_k u_{k-1}}.$$
In particular I would like to find $\lim_{k\rightarrow \infty} u_k$ for this sequence.
For a value $\bar u$ to be a potential limiting point, this value should at least satisfy:
$$
\bar u = \frac{1}{2-\bar u^2},
$$
it is not hard to see that this equation has $3$ solutions, namely:
$$
\bar u = 1, -\frac{1}{2}(1+\sqrt{5}), \frac{1}{2}(-1+\sqrt{5}).
$$
I can numerically verify that the first two candidates do not attract the sequence $(u_k)_k$, thus we only have $u_k \rightarrow 1$ resp. $u_k\rightarrow -\frac{1}{2}(1+\sqrt{5})$ when $u_0,u_1 = 1$ resp. $u_0,u_1=-\frac{1}{2}(1+\sqrt{5})$ whilst the third seems to be an attractor for the sequence.
Furthermore there is the special case where $u_0 \cdot u_1 = 2$, which we exclude as $u_2$ is not defined in this case.
However, I am unable to prove that for any $u_0, u_1 \notin \{-1, -\frac{1}{2}(1+\sqrt{5})\}$ and $u_0 \cdot u_1 \neq 2$, we have $u_k \rightarrow \frac{1}{2} (-1+\sqrt{5})$.
A related problem was studied here, and I understand the proof given there, but I do not see how to extend this new problem.
 A: We are lucky : one can find a closed-form expression for $u_n$ (see (1) below). This will allow us to show that  the sequence always converges and to determine the limit.
Let us compute a first few terms. Putting $x=u_1,y=u_0$ we have $u_2=\frac{1}{2-xy}$, $u_3=\frac{2-xy}{4-x-2xy}$, $u_4=\frac{4-x-2xy}{7-2x-4xy}$, $u_5=\frac{7-2x-4xy}{12-4x-7xy}$, $u_6=\frac{12-4x-7xy}{20-7x-12xy}$.
We start to see a pattern here. First, the denominator of $u_n$ seems to coincide with the numerator of $u_{n+1}$, and the coefficients $1,2,4,7,12,20$
seem to be the Fibonacci numbers minus 1. To summarize, the following holds for
$n\leq 6$ :
$$
u_n=\frac{v_n}{v_{n+1}}, \ \textrm{where} \ v_n=(F_{n+1}-1)-(F_{n-1}-1)x-(F_{n}-1)xy \tag{1}
$$
Now that we have guessed (1), we can prove it rigorously by induction : if (1) holds for $n$ and $n+1$, we have
$$u_{n+2}=\frac{1}{2-u_nu_{n+1}}=\frac{1}{2-\frac{v_n}{v_{n+2}}}=\frac{v_{n+2}}{2v_{n+2}-v_n} \tag{2}$$
Notice that the Fibonacci sequence satisfies
$$2F_{n+2}-F_n=F_{n+2}+(F_{n+2}-F_n)=F_{n+2}+F_{n+1}=F_{n+3} \tag{3}$$
Since $(v_n)$ is a linear combination of $F_{n-1},F_n$ and $F_{n+1}$, it will satisfy this linear recurrence as well : $2v_{n+2}-v_n=v_{n+3}$, so that (2) gives $u_{n+2}=\frac{v_{n+2}}{v_{n+3}}$ which finishes the proof of (1) by induction.
Let $\phi$ be the golden ratio,  $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=1-\phi=\frac{1-\sqrt{5}}{2}$ is the conjugate of $\phi$. Using the well-known formula $F_n=\frac{\phi^n-\psi^n}{\sqrt{5}}$, we deduce
$$
v_n=\frac{\phi^2-x-\phi xy}{\sqrt{5}}\phi^{n-1}-\frac{\psi^2-x-\psi xy}{\sqrt{5}}\psi^{n-1}+(xy+x-1)\tag{4}
$$
There are now several cases to consider. If $c_1=\frac{\phi^2-x-\phi xy}{\sqrt{5}}$ is nonzero, then $v_n \sim c_1 \phi^{n-1}$ as $n\to\infty$, so that $u_n$ converges to $\phi$.
Next, assume that $c_1=0$ but $c_2=xy+x-1$ is nonzero, then $v_{n} \to c_2 $ as $n\to\infty$, so that $(u_n)$ converges to $1$.
Finally, if both $c_1$ and $c_2$ are zero, then $x=y=1$ and $(u_n)$ is constant equal to $1$.
